Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar. Suppose you prepare a glucose solution using the described procedure. Step 1: Dissolve 259.2 g of glucose in enough water to make 500.0 mL of solution. Step 2: Transfer 20.4 mL of the solution to a new flask and add enough water to make 250.0 mL of dilute solution. What is the concentration (in M) of the glucose solution at the end of the procedure?

Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar. Suppose you prepare a glucose solution using the described procedure. Step 1: Dissolve 259.2 g of glucose in enough water to make 500.0 mL of solution. Step 2: Transfer 20.4 mL of the solution to a new flask and add enough water to make 250.0 mL of dilute solution. What is the concentration (in M) of the glucose solution at the end of the procedure?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter3: Mass Relations In Chemistry; Stoichiometry

Section: Chapter Questions

Problem 23QAP

See similar textbooks

Similar questions

What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
A soft drink contains an unknown mass of citric acid, C3H5O(COOH)3. It requires 6.42 mL of 9.580 × 10−2-M NaOH to neutralize the citric acid in 10.0 mL of the soft drink. C3H5O(COOH)3(aq) + 3 NaOH(aq) → Na3C3H5O(COO)3(aq) + 3 H2O(ℓ) Determine which step in these calculations for the mass of citric acid in 1 mL soft drink is incorrect? Why? n (NaOH) = (6.42 mL)(1L/1000 mL)(9.580 × 10−2 mol/L) n (citric acid) = (6.15 × 10−4 mol NaOH) × (3 mol citric acid/1 mol NaOH) m (citric acid in sample) = (1.85 × 10−3 mol citric acid) × (192.12 g/mol citric acid) m (citric acid in 1 mL soft drink) = (0.354 g citric acid)/(10 mL soft drink) Determine the correct result.
You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic-molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.
What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm-3 and contains 37.21% HCl by mass?
You want to prepare a 1.0 mol/kg solution of ethyleneglycol, C2H4(OH)2, in water. Calculate the mass of ethylene glycol you would need to mix with 950. g water.
A student weighs out a 4.80-g sample of aluminum bromide, transfers it to a 100-mL volumetric flask, adds enough water to dissolve it, and then adds water to the 100-mL mark. What is the molarity of aluminum bromide in the resulting solution?
What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?
Magnesium sulfate (MgSO4) has a solubility of 38.9 g/ 100 g H2O at 30C. A solution is prepared by adding 9.50 g of MgSO4 to 25.0 g of water at 40C. A homogeneous mixture is obtained. Is the solution saturated, unsaturated, or supersaturated? One gram of magnesium sulfate is added to the solution cooled to 30C. Would you expect some of the MgSO4 to precipitate? If so, how much? If not, how much more MgSO4 can be added before precipitation takes place?
The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 106 parts of solution. Mathematically, by mass: ppm=gsolutegsolution=mgsolutekgsolution In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0 g of solute per 1.0 mL, which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 ppb Hg in H2O b. 1.0 ppb CHCl3 in H2O c. 10.0 ppm As in H2O d. 0.10 ppm DDT (C14H9Cl5) in H2O